Sum of Reciprocals of Squares of Odd Integers/Proof 2
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Theorem
| \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) | \(=\) | \(\ds 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\pi^2} 8\) |
Proof
Let $n$ be a positive integer.
| \(\ds \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \rd x\) | \(=\) | \(\ds \int_0^{\pi / 2} \sin^{2 n + 1} x \rd x\) | substituting $x \to \sin x$ | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac {2^{2 n} \paren {n!}^2} {\paren {2 n + 1}!}\) | Reduction Formula for Definite Integral of Power of Sine |
We have:
| \(\ds \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \rd x\) | \(=\) | \(\ds \int_0^1 \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \rd x\) | Power Series Expansion for Real Arcsine Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2} } \rd x\) | Interchange of sum and integral is valid by Tonelli's Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n + 1} } \frac {2^{2 n} \paren {n!}^2} {\paren {2 n + 1}!}\) | by $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2}\) |
We also have:
| \(\ds \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2} } \rd x\) | \(=\) | \(\ds \intlimits {\frac 1 2 \paren {\arcsin x}^2} 0 1\) | Fundamental Theorem of Calculus | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\pi^2} 8\) |
So we can deduce:
- $\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} = \frac {\pi^2} 8$
$\blacksquare$